Strength of materials—bars under self-weight: What is the ratio of the elongation of a conical bar due to its own weight to that of a prismatic bar of the same material and length (both hanging vertically)?

Introduction / Context:
Bars hanging under their own weight elongate because the internal axial force increases toward the support. The distribution of cross-sectional area changes the strain distribution. This problem compares a conical (linearly tapered) bar to a prismatic (constant area) bar, a classical result in strength of materials.

Given Data / Assumptions:

• Both bars have the same length L and material properties (density rho and modulus E).
• Bars hang vertically; weight density is rho * g.
• Conical bar area varies as A(y) = A0 * (1 - y/L)^2, y from the top.
• Self-weight only; no external tip load.

Concept / Approach:
Total elongation is the integral of axial strain along the length. At a distance y from the top, axial force equals the weight of the portion below: W(y) = rho * g * ∫(y to L) A(s) ds. Strain(y) = stress/E = [W(y) / A(y)] / E. Integrate strain from 0 to L.

Step-by-Step Solution:
Prismatic bar: A(y) = A0.Force at y: Wp(y) = rho * g * A0 * (L - y).Strain: ep(y) = [rho * g * (L - y)] / E.Elongation: δp = ∫(0→L) ep(y) dy = (rho * g / E) * [L^2 / 2] = rho * g * L^2 / (2E).Conical bar: A(y) = A0 * (1 - y/L)^2.Weight below y: Wc(y) = rho * g * ∫(y→L) A0 * (1 - s/L)^2 ds.Strain: ec(y) = [Wc(y) / A(y)] / E.Integrating yields δc = rho * g * L^2 / (6E).Ratio δc / δp = (1/6) / (1/2) = 1/3.

Verification / Alternative check:
Since the conical bar is thicker near the support (where forces are higher), it stretches less than a prismatic bar; a ratio less than 1 is expected. 1/3 matches this intuition.

Why Other Options Are Wrong:
1/2, 2/3, 3/2, and 1 contradict the derived integral result and the physical trend.

Common Pitfalls:
Using average weight or confusing area variation; forgetting that self-weight force varies along the length and must be integrated.

Final Answer:
1/3